[tex]\large\boxed{\begin{array}{l}\rm Sendo\,c\,e\,l\,o\,comprimento\,e\,a\,largura\\\rm do\,ret\hat angulo\\\rm podemos\,usar\,o\,teorema\,de\,pit\acute agoras\\\rm para\,encontrar\,a\,outra\,dimens\tilde ao\,(fig\,1)\end{array}}[/tex]
[tex]\large\boxed{\begin{array}{l}\rm \ell^2+8^2=10^2\\\rm\ell^2+64=100\\\rm\ell ^2=100-64\\\rm \ell^2=36\\\rm\ell=\sqrt{36}\\\rm\ell=6\,dam\end{array}}[/tex]
[tex]\large\boxed{\begin{array}{l}\rm para\,encontrar\,o\,per\acute imetro,basta\\\rm somar\,as\,dimens\tilde oes\,e\,multiplicar\,por\,2\\\rm pois\,o\,mesmo\,possui\,lados\,iguais\\\rm dois\,a\,dois.\\\rm Dessa,forma\,teremos:\\\rm p=2\cdot(c+\ell)\\\rm p=2\cdot(8+6)\\\rm p=2\cdot 14\\\rm p=28~dam\end{array}}[/tex]
